Asked by Sam
A 10.3g sample of an elemental gas has a volume of 58.4L at 758torr when the temperature is 2.5 C. Determine the identity of the gas.
Answers
Answered by
Damon
we are close to STP here where one mol is 22.4 liters
58.4/22.4 is
around 2.61 moles
10.3/2.61 = 3.95 grams/mol
Helium maybe ?
58.4/22.4 is
around 2.61 moles
10.3/2.61 = 3.95 grams/mol
Helium maybe ?
Answered by
DrBob222
Damon's estimate is quite good (and reasonable) and the answer is correct. If you want to do it without assuming it is 22.4 L/mol, it is done this way.
Use PV = nRT and solve for n = number of mols of the gas.
(758*58.4)/(760*0.08206*275.6) = 0.257 = n.
Then mols = grams/molar mass or
molar mass = g/mol = 10.3/0.257 = 4.01. Indeed that is He.
Use PV = nRT and solve for n = number of mols of the gas.
(758*58.4)/(760*0.08206*275.6) = 0.257 = n.
Then mols = grams/molar mass or
molar mass = g/mol = 10.3/0.257 = 4.01. Indeed that is He.
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